In a single slit diffraction pattern,the wavelength of light used is $628 \text{ nm}$ and slit width is $0.2 \text{ mm}$. The angular width of the central maximum is $\alpha \times 10^{-2} \text{ degrees}$. The value of $\alpha$ is . . . . . . . (Take $\pi = 3.14$)

$A$ monochromatic light of wavelength $6000 \text{ Å}$ is incident on a single slit of width $0.01 \text{ mm}$. If the diffraction pattern is formed at the focus of a convex lens of focal length $20 \text{ cm}$, the linear width of the central maximum is: (in $\text{ mm}$)

$A$ light source with a wavelength of $5000 \, \mathring A$ produces a single-slit diffraction pattern. The first minimum in the diffraction pattern is observed at a distance of $5 \, mm$ from the central maximum. The distance between the slit and the screen is $2 \, m$. Find the width of the slit. (in $, mm$)

$A$ single slit diffraction pattern is formed with light of wavelength $6195 Å$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength $\lambda_0$. The value of $\lambda_0$ is (in $Å$)

The diffraction effect can be observed in

In a single-slit diffraction experiment,what is the width of the slit $(b)$ such that $20$ maxima of a double-slit interference pattern are obtained within the central maximum of the diffraction pattern? (Given: Slit separation for the double-slit arrangement $d = 2 \, mm$)